Consider a TCP connection between hosts A and B. At some time the advertised win

Question

Consider a TCP connection between hosts A and B. At some time the advertised window size from B to A is 2 MSS. Every packet transmitted from A to B is successful with probabiliity p and is dropped with probability 1-p whereas the transmission from B to A is error-free. What is the total expected time for A to successfully transmit 2 packets? The time for successful transmission is measured from the moment of the transmission of the first packet until the reception of ACK for the second packet. Assume propagation delay between A and B is given by T. Ignore transmission delay and processing time at each host. Assume the retransmission timeout value as 4T . Also assume that the retransmission is triggered only by timeouts but not duplicate ACKs

Explanation / Answer

Answer: Total expected time to successfully transmit a packet/segment from host A to host B form the momemt of starting transmission of packet till ACK is received is referred to as Round Trip Time (RTT) defined as below:

RTT = 2*Time required to deliver the packet + processing time.

For the given problem, processing time can be ignored. Hence,

RTT = 2*Time required to deliver the packet

where Time required to deliver the packet is given as:

Time required to deliver the packet = Transmission time + Propagation delay + Queueing time

For the given problem, Transmission time and Queueing time can be ignored, while Propagation delay between A and B is T.

Hence,

Time required to deliver the packet = T

Hence,

RTT = 2T

As per the problem, 2 packets are to be sent between A and B. Since, window size is 2 MSS, hence both the packets can be sent in the same transmission. Therefore, overall expected time will be

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